What is meant here is this. In ΔABC, angle trisectors are drawn that form Morley's triangle A'B'C'. Extend the trisectors till their intersection with the circumcircle of ΔABC. AC' cuts it at A1, AB' at A2, and so on. The lines A1B2, B1C2 and C1A2 form ΔA0B0C0, which is equilateral.

An easy proof follows from the observation that lines A1B2, B1C2 and C1A2 are parallel to the sides of Morley's triangle.

For example, we know that ∠BA'C' = C/3 + 60°. Denote the point of intersection of C1A2 and BB1 as P. I am going to show that ∠BPC1 = C/3 + 60°, from which C1A2||A'C', and similarly for other two sides.